How to create an Identity matrix and replace the ''1''s with n which n is the elements' row or column number?

Question

I want to create a n by n Identity matrix and replace the ones with n π^2. NOS is three here (it is a number). The code I have written for this doesn't work.

Kin = IdentityMatrix[NOS]

n = 1; m = 1; While[n == m && n < NOS &&  m < NOS, 
 Kin[[n, m]] -> n π^2; n++; m++]

I want

Kin[[1, 1]]== π^2
Kin[[2, 2]]==2 π^2

and so on. Could anyone help me with this please?

Answer 1

It is far faster to make the multiplication by a vector and then convert to a diagonal matrix, rather than building a full matrix and multiplying after:

DiagonalMatrix[π^2 Range[3]]

$\left( \begin{array}{ccc} \pi ^2 & 0 & 0 \\ 0 & 2 \pi ^2 & 0 \\ 0 & 0 & 3 \pi ^2 \\ \end{array} \right)$

Even faster is to build a sparse array, which is done by applying SparseArray before (inside) DiagonalMatrix. With this formulation multiplicaiton after is fast because only specified elements (the diagonal) are actually multiplied.

π^2 DiagonalMatrix[Range[1000]];                // RepeatedTiming // First
DiagonalMatrix[π^2 Range[1000]];                // RepeatedTiming // First
π^2 DiagonalMatrix[SparseArray @ Range[1000]];  // RepeatedTiming // First

0.434

0.01

0.000647

The examples above preserve the diagonals in Symbolic form. If machine reals⁽¹⁾ are OK:

π^2` DiagonalMatrix[Range[1000]];                // RepeatedTiming // First
DiagonalMatrix[π^2` Range[1000]];                // RepeatedTiming // First
π^2` DiagonalMatrix[SparseArray @ Range[1000]];  // RepeatedTiming // First

0.0040

0.00174

0.0000231

Reference:

• What is a Mathematica packed array?

Answer 2

NOS = 3; 

π^2 DiagonalMatrix[Range[NOS]] 
π^2 Range[NOS] IdentityMatrix[NOS]
MapIndexed[π^2 #2[[1]] # &, IdentityMatrix[NOS], {2}]
ReplacePart[IdentityMatrix[NOS], {i_, i_} :> (i π^2)]
Block[{i = 1}, IdentityMatrix[NOS] /. 1 :> (π^2 i++)]

all give

{{π^2, 0, 0}, {0, 2 π^2, 0}, {0, 0, 3 π^2}}

Answer 3

here is another way using Unitize

π^2 Table[i (1 - Unitize[Range[1000] - i]), {i, 1000}]; // AbsoluteTiming 
  (* {0.552622, Null} *)

and with SparseArray alone:

π^2 SparseArray[{i_, i_} :> i , {1000, 1000}];//AbsoluteTiming
  (* {0.00134081, Null} *)

Borrowing the backtick from @Mr.Wizard significantly improves the answer

π^2` Table[i (1 - Unitize[Range[1000] - i]), {i, 1000}]; // AbsoluteTiming
(* {0.0230811, Null} *)


π^2` SparseArray[{i_, i_} :> i, {1000, 1000}]; // AbsoluteTiming
(* {0.000165618, Null} *)